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Unformatted text preview: homework 09 – LEE, BENJAMIN – Due: Sep 25 2007, 4:00 am 1 Question 1, chap 26, sect 1. part 1 of 1 10 points Two large parallel conducting plates P and Q are connected to a battery of emf E , as shown. A test charge is placed successively at points I , II , and III . E +- I II III If edge effects are negligible, the force on the charge when it is at point III is 1. equal in magnitude to the force on the charge when it is at point I , but in the oppo- site direction. 2. of equal magnitude and in the same di- rection as the force on the charge when it is at point I . 3. much greater in magnitude than the force on the charge when it is at point II , but in the same direction. 4. much less in magnitude than the force on the charge when it is at point II , but in the same direction. 5. of equal magnitude and in the same di- rection as the force on the charge when it is at point II . correct Explanation: Neglecting edge effects, the electric field strength between the two plates is uniform. Therefore the force on the test charge is the same at points II and III , both in magnitude and in direction. When edge effects are negligible, the elec- tric field strength at point I is zero, so the force on the test charge is zero when it is at point I . Question 2, chap 26, sect 1. part 1 of 1 10 points A parallel plate capacitor is connected to a battery. + Q- Q d b b 2 d b b If we double the plate separation, 1. the capacitance is doubled. 2. the potential difference is halved. 3. the electric field is doubled. 4. None of these. 5. the charge on each plate is halved. cor- rect Explanation: The capacitance of a parallel plate capaci- tor is C = ǫ A d . Hence doubling d halves the capacitance, and Q = C V is also halved parenleftbigg C ′ = ǫ A 2 d = 1 2 ǫ A d = 1 2 C parenrightbigg . Question 3, chap 26, sect 1....
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